We know that
a(n)=4n+1
To find the sum of the first n<span> terms of an arithmetic series
use the formula
</span><span><span>Sn</span>=<span>n(<span>a1</span> + <span>an</span>)/2
</span></span>a1--------------> is the first term
an--------------> is the last term
n--------------- > <span>is the number of terms
</span>we have that
a1------------> 4*(1)+1=5
a30----------> 4*(30)+1=121
n------------- > 30
S30=30*(5 + 121)/2=1890
the answer is 1890
Step-by-step explanation:
The vertex form of the equation for a parabola is given by
where (h, k) are the coordinates of the parabola's vertex. Since the vertex is at (1, -6), we can write the equation as
Also, since the parabola passes through (4, -7), we can use this to find the value for a:
or
Therefore, the equation of the parabola is
Assuming the question marks are minus signs
to find max, take derivitive and test 0's and endpoints
take derivitive
f'(x)=18x²-18x-108
it equal 0 at x=-2 and 3
if we make a sign chart to find the change of signs
the sign changes from (+) to (-) at x=-2 and from (-) to (+) at x=3
so a reletive max at x=-2 and a reletive min at x=3
test entpoints
f(-3)=83
f(-2)=134
f(3)=-241
f(4)=-190
the min is at x=3 and max is at x=-2
Step-by-step explanation:
Consider the provided information.
Part (A) We need to find the confidence coefficient.
The given value or percentage of probability is known as the confidence coefficient that the interval contains the parameter.
Hence, the confidence coefficient is 0.90
Part (B) Practical interpretation of both of the 99% confidence intervals.
It means that there is 99% confident that the mean of population HRV for officers diagnosed hypertension which lies between 6.9 and 122.9
There is 99% confident that the mean of population HRV for officers those are not hypertensive which lies between 144.2 and 184.9.
Part (C)
If you are saying 99% confident that means the 99% of the similarly generated confidence intervals will contain the true value of the population mean in repeated sampling.
Part (D)
If We want to reduce the width of each confidence interval, you need to use smaller confidence coefficient.